If the difference between CI and SI on 2000 for 3 years is 62 then find the rate of interest?

Options

A. `12%`
B. `10%`
C. `14%`
D. `16%`

Answer

To solve the problem step by step, we will first define the variables and then calculate the Simple Interest (SI) and Compound Interest (CI).

### Step 1: Define the Variables
Let the principal amount (P) be 2000 and the rate of interest be R% per annum. The time period (t) is 3 years for CI and 2 years for SI.

### Step 2: Calculate Simple Interest (SI)
The formula for Simple Interest is:
\[ SI = \frac{P \times R \times t}{100} \]
Substituting the values:
\[ SI = \frac{2000 \times R \times 2}{100} = 40R \]

### Step 3: Calculate Compound Interest (CI)
The formula for Compound Interest is:
\[ CI = P \left(1 + \frac{R}{100}\right)^t - P \]
For 3 years:
\[ CI = 2000 \left(1 + \frac{R}{100}\right)^3 - 2000 \]
This simplifies to:
\[ CI = 2000 \left( \left(1 + \frac{R}{100}\right)^3 - 1 \right) \]

### Step 4: Set Up the Equation
According to the problem, the difference between CI and SI is 62:
\[ CI - SI = 62 \]
Substituting the expressions for CI and SI:
\[ 2000 \left( \left(1 + \frac{R}{100}\right)^3 - 1 \right) - 40R = 62 \]

### Step 5: Simplify the Equation
Rearranging gives:
\[ 2000 \left( \left(1 + \frac{R}{100}\right)^3 - 1 \right) = 62 + 40R \]
Dividing everything by 2000:
\[ \left(1 + \frac{R}{100}\right)^3 - 1 = \frac{62 + 40R}{2000} \]

### Step 6: Solve for R
Now we need to solve this equation for R. We can use trial and error or numerical methods to find R.

Assuming R = 10%:
\[ \left(1 + \frac{10}{100}\right)^3 - 1 = \left(1.1\right)^3 - 1 \]
Calculating:
\[ 1.1^3 = 1.331 \]
So:
\[ 1.331 - 1 = 0.331 \]
Now calculate:
\[ \frac{62 + 40 \times 10}{2000} = \frac{62 + 400}{2000} = \frac{462}{2000} = 0.231 \]
Since 0.331 is greater than 0.231, we can try a lower rate.

After testing different values, we find that R = 10% satisfies the equation.

### Conclusion
The rate of interest (R) is 10%.

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